Math2.org Math Tables: Derivatives of Trig Functions
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| (Math) |
 sin(x) = cos(x) cos(x) = -sin(x) tan(x) = sec2(x) csc(x) = -csc(x) cot(x) sec(x) = sec(x) tan(x) cot(x) = -csc2(x)
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sin(x)
: algebraic Method
Given: lim(d->0) sin(d)/d = 1.
Solve:
= lim ( sin(x)cos(d) + cos(x)sin(d) - sin(x) ) / d
= lim ( sin(x)cos(d) - sin(x) )/d + lim cos(x)sin(d)/d
= sin(x) lim ( cos(d) - 1 )/d + cos(x) lim sin(d)/d
= sin(x) lim ( (cos(d)-1)(cos(d)+1) ) / ( d(cos(d)+1) ) + cos(x) lim sin(d)/d
= sin(x) lim ( cos2(d)-1 ) / ( d(cos(d)+1 ) + cos(x) lim sin(d)/d
= sin(x) lim -sin2(d) / ( d(cos(d) + 1) + cos(x) lim sin(d)/d
= sin(x) lim (-sin(d)) * lim sin(d)/d * lim 1/(cos(d)+1) + cos(x) lim sin(d)/d
= sin(x) * 0 * 1 * 1/2 + cos(x) * 1 = cos(x) Q.E.D.
Proof of cos(x)
: from the derivative of sine
This can be derived just like
sin(x) was derived or more easily from the result of
sin(x)
Given:
sin(x) = cos(x); Chain Rule.
Solve:
cos(x) = sin(x + PI/2)
=sin(u) *
= cos(u) * 1 = cos(x + PI/2) = -sin(x) Q.E.D.
Proof of tan(x)
: from the derivatives of sine and cosine
Given:
sin(x) = cos(x);
cos(x) = -sin(x); Quotient Rule.
Solve:
tan(x) = sin(x) / cos(x)
= ( cos(x)
= ( cos(x)cos(x) + sin(x)sin(x) ) / cos2(x)
= 1 + tan2(x) = sec2(x) Q.E.D.
Proof of csc(x),
sec(x),
cot(x)
: from derivatives of their reciprocal functions
Given:
sin(x) = cos(x);
cos(x) = -sin(x);
tan(x) = sec2(x); Quotient Rule.
Solve:
